EMIS 8374 Max-Flow in Undirected Networks Updated 18 March 2008

Max Flow in Undirected Networks 1 4 10 3 8 6 5 5 2 s t 6 2 10 3 4 1 4

Replace Edge {i,j} With Arcs (i,j) and (j,i) 4 6 4 1 t 6 5 5 5 10 2 10 10 2 10 6 6 3 3 1 1 3 s 4 2 4 4 8 8

Max Flow in Directed Network 4 1 t 6 2 4 4 2 10 10 6 3 1 3 s 4 2 4 7

Max Flow in Directed Network 6 4 4 1 2 t 4 10 10 6 3 3 2 1 4 7 s 4

Remove Bi-directional flows if xij  xji then xij = xij – xji and xji = 0 else xji = xji – xij and xij = 0 1 2 4 6 12 10 1 4 6 2-2=0 4-2=2 2 12 10

Max Flow in Undirected Network Arrows indicate flow direction 1 4 6 2 2 4 10 s t 6 2 10 3 3 4 7 1 4

Remove Saturated Edges 1 2 s t 3 S = {s, 4} T = {1, 2, 4, t} 4

Undirected s-t Cut 1 4 6 2 2 4 10 s t 6 2 10 3 3 4 7 1 u[S, T] = 24 4

All-Pairs Minimum Cut Problem Find the minimum value of u[A, B] where [A, B] is an partition of the nodes such that |A|>0 and |B|>0. Also known as the minimum 2-cut. Note that no specific source or sink nodes are specified.

Min 2-Cut Algorithm Since the network is undirected, u[A, B] = u[A, B] Don’t need to try s = j and t = i if we’ve already tried s = i and t = j

Min 2-Cut Algorithm v* = ; for s = 1 .. |N| - 1 for t = s + 1 .. |N| begin solve max s-t flow problem; identify min cut [S, T]; if u[S, T] < v* then A = S; B = T; end

Min 2-Cut Example 10 1 2 3 3 3 1 8 5 3 4 2 5 4

Minimum Cut: s = 1, t = 2 10 1 2 3 3 3 1 8 5 3 4 2 5 S = {1} T = {2, 3, 4, 5} u[S, T] = 17 4

Minimum Cut s = 1, t = 3 1 2 4 3 5 10 8 S = {1,2} T = {3,4,5} u[S, T]=14

Minimum Cut: s = 1, t = 4 10 1 2 3 3 3 1 5 8 3 4 2 5 S = {1,2,3,5} T={4} u[S, T]=12 4

Minimum Cut: s = 1, t = 5 1 2 4 3 5 10 8 S = {1,2} T = {3,4,5} u[S, T]=14

Minimum Cut: s = 2, t =3 10 1 2 3 3 3 1 5 8 3 4 2 5 S = {2,1} T = {3,4,5} u[S, T]=14 4

Observation Suppose s = 2 and t = 3 and let [A, B] be a minimum 2-3 cut. Case 1: node 1 is in A [A, B] is also a 1-3 cut Thus, we already know u[A, B]  14 Case 2: node 1 is in N2 [A, B] is also a 2-1 (1-2) cut Thus, we already know u[A, B]  17 There is no need to solve the max-flow problem for s = 2 and t =3.

An Improved Min 2-Cut Algorithm Consider a minimum 2-cut [A, B] Let A be the set containing node 1. Since |B| > 0, it must contain at least one node in {2, 3, 4, 5}. Thus we can discover [A, B] by solving only |N| - 1 max flow problems with s =1 and t = 2, t = 3, …, t = |N|. Complexity is O(n f(n, m)) where f(n, m) is the complexity of solving a max flow problem

Minimum 2-Cut 10 1 2 3 3 3 1 5 8 3 4 2 5 A = {1,2,3,5} B = {4} 4 u[A, B]=12

Edge Connectivity In a so-called unweighted graph where each edge as a capacity of 1 unit, the capacity of a minimum 2-cut is known as the edge connectivity of the graph Connectivity is an important measure of a network’s reliability. In a telecommunications network an edge connectivity of two (2) means that the network can survive single-link failures.